Search results
Results from the WOW.Com Content Network
Substituting into the Clapeyron equation =, we can obtain the Clausius–Clapeyron equation [8]: 509 = for low temperatures and pressures, [8]: 509 where is the specific latent heat of the substance. Instead of the specific, corresponding molar values (i.e. L {\\displaystyle L} in kJ/mol and R = 8.31 J/(mol⋅K)) may also be used.
Isotherms of an ideal gas for different temperatures. The curved lines are rectangular hyperbolae of the form y = a/x. They represent the relationship between pressure (on the vertical axis) and volume (on the horizontal axis) for an ideal gas at different temperatures: lines that are farther away from the origin (that is, lines that are nearer to the top right-hand corner of the diagram ...
An example of a system that does not have a unique ground state is one whose net spin is a half-integer, for which time-reversal symmetry gives two degenerate ground states. For such systems, the entropy at zero temperature is at least k B ln(2) (which is negligible on a macroscopic scale).
The extent of boiling-point elevation can be calculated by applying Clausius–Clapeyron relation and Raoult's law together with the assumption of the non-volatility of the solute. The result is that in dilute ideal solutions, the extent of boiling-point elevation is directly proportional to the molal concentration (amount of substance per mass ...
The German physicist Rudolf Clausius learned of Carnot's work through Clapeyron's memoir. Clausius corrected Carnot's theory by replacing the conservation of caloric with the work-heat equivalence (i.e., energy conservation). Clausius also put the second law of thermodynamics into mathematical form by defining the concept of entropy.
In the linear theory of elasticity Clapeyron's theorem states that the potential energy of deformation of a body, which is in equilibrium under a given load, is equal to half the work done by the external forces computed assuming these forces had remained constant from the initial state to the final state.
This result seems to contradict the equation dF = −S dT − P dV, as keeping T and V constant seems to imply dF = 0, and hence F = constant. In reality there is no contradiction: In a simple one-component system, to which the validity of the equation d F = − S d T − P d V is restricted, no process can occur at constant T and V , since ...
If one sets out to determine the specific volume of an ideal gas, such as super heated steam, using the equation ν = RT/P, where pressure is 2500 lbf/in 2, R is 0.596, temperature is 1960 °R. In that case, the specific volume would equal 0.4672 in 3 /lb.